Question

[Trần Lê Hải Yến]

Cho M =5/3+8/3^2+11/3^3+.....+302/3^100.CMR M<hỗn số 3 và 1/2

Answer 1

[Dinhhuy]

Ta có: \(M=\frac53+\frac{8}{3^2}+\frac{11}{3^3}+\cdots+\frac{302}{3^{100}}\)

\(\to3M=3\left(\frac53+\frac{8}{3^2}+\frac{11}{3^3}+\cdots+\frac{302}{3^{100}}\right)\)

\(\to3M-M=\left(5+\frac83+\frac{11}{3^2}+\cdots+\frac{302}{3^{100}}\right)-\left(\frac53+\frac{8}{3^2}+\frac{11}{3^3}+\cdots+\frac{302}{3^{100}}\right)\)

\(\to2M=5+1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}-\frac{302}{3^{100}}\left(1\right)\)

Đặt \(A=\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\)

\(3A=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{97}}\)

\(3A-A=\left(1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{97}}\right)-\left(\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\right)\)

\(2A=1-\frac{1}{3^{98}}\)

\(\to A=\frac{1-\frac{1}{3^{98}}}{2}\left(2\right)\)

Thay \(\left(2\right)\to\left(1\right)\) , ta có:

\(\to2M=5+1+\frac{1-3^{98}}{2}-\frac{302}{3^{100}}\)

\(\to M=\frac62+\frac{1-3^{98}}{2}-\frac{302}{3^{100}}\)

\(\to M=3+\frac{1-3^{98}}{2}-\frac{302}{3^{100}}\)

Mà: \(\frac{1-3^{98}}{2}-\frac{302}{3^{100}}<\frac12\)

Vậy \(M<3\frac12\) (đpcm)